79886
The points of discontinuity of the function \(\begin{aligned} & f(x)=\frac{1}{x-1} \text { if } 0 \leq x \leq 2 \\ & f(x)=\frac{x+5}{x+3} \text { if } 2\lt x \leq 4\end{aligned}\), in its domain are
1 \(x=2\) only
2 \(x=1, x=2\)
3 \(x=0, x=2\)
4 \(x=4\) only
Explanation:
(B) : Given, \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}-1}\) if \(0 \leq \mathrm{x} \leq 2\) at \(x=1=\lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1} \frac{1}{x-1}=\frac{1}{0}=\infty\) \(\therefore \mathrm{f}(\mathrm{x})\) is discontinuous at \(\mathrm{x}=1\) And, \(f(x)=\frac{x+5}{x+3}\) if \(2\lt x \leq 4\) at \(\mathrm{x}=2\) \(\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}} \frac{x+5}{x+3}=\frac{2+5}{2+3}=\frac{7}{5}\) \(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} \frac{1}{x-1}=\frac{1}{2-1}=1\) \(\therefore \lim _{x \rightarrow 2^{+}} \mathrm{f}(\mathrm{x}) \neq \lim _{\mathrm{x} \rightarrow 2^{-}} \mathrm{f}(\mathrm{x})\) \(\therefore \mathrm{f}(\mathrm{x})\) is discontinuous at \(\mathrm{x}=2\)
MHT CET-2020
Limits, Continuity and Differentiability
79887
If the function \(f(x)=\frac{(1-\cos 5 x) \sin 5 x}{x^{2} \sin 3 x}, x \neq 0\), is continuous at \(x=0\), then \(f(0)=\)
79889
If \(\begin{aligned} f(x) & =6 \beta-3 \alpha x, \text { if }-4 \leq x\lt -2 \\ & =4 x+1, \text { if }-2 \leq x \leq 2\end{aligned}\) is continuous on \([-4,2]\), then \(\alpha+\beta=\)
79886
The points of discontinuity of the function \(\begin{aligned} & f(x)=\frac{1}{x-1} \text { if } 0 \leq x \leq 2 \\ & f(x)=\frac{x+5}{x+3} \text { if } 2\lt x \leq 4\end{aligned}\), in its domain are
1 \(x=2\) only
2 \(x=1, x=2\)
3 \(x=0, x=2\)
4 \(x=4\) only
Explanation:
(B) : Given, \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}-1}\) if \(0 \leq \mathrm{x} \leq 2\) at \(x=1=\lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1} \frac{1}{x-1}=\frac{1}{0}=\infty\) \(\therefore \mathrm{f}(\mathrm{x})\) is discontinuous at \(\mathrm{x}=1\) And, \(f(x)=\frac{x+5}{x+3}\) if \(2\lt x \leq 4\) at \(\mathrm{x}=2\) \(\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}} \frac{x+5}{x+3}=\frac{2+5}{2+3}=\frac{7}{5}\) \(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} \frac{1}{x-1}=\frac{1}{2-1}=1\) \(\therefore \lim _{x \rightarrow 2^{+}} \mathrm{f}(\mathrm{x}) \neq \lim _{\mathrm{x} \rightarrow 2^{-}} \mathrm{f}(\mathrm{x})\) \(\therefore \mathrm{f}(\mathrm{x})\) is discontinuous at \(\mathrm{x}=2\)
MHT CET-2020
Limits, Continuity and Differentiability
79887
If the function \(f(x)=\frac{(1-\cos 5 x) \sin 5 x}{x^{2} \sin 3 x}, x \neq 0\), is continuous at \(x=0\), then \(f(0)=\)
79889
If \(\begin{aligned} f(x) & =6 \beta-3 \alpha x, \text { if }-4 \leq x\lt -2 \\ & =4 x+1, \text { if }-2 \leq x \leq 2\end{aligned}\) is continuous on \([-4,2]\), then \(\alpha+\beta=\)
79886
The points of discontinuity of the function \(\begin{aligned} & f(x)=\frac{1}{x-1} \text { if } 0 \leq x \leq 2 \\ & f(x)=\frac{x+5}{x+3} \text { if } 2\lt x \leq 4\end{aligned}\), in its domain are
1 \(x=2\) only
2 \(x=1, x=2\)
3 \(x=0, x=2\)
4 \(x=4\) only
Explanation:
(B) : Given, \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}-1}\) if \(0 \leq \mathrm{x} \leq 2\) at \(x=1=\lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1} \frac{1}{x-1}=\frac{1}{0}=\infty\) \(\therefore \mathrm{f}(\mathrm{x})\) is discontinuous at \(\mathrm{x}=1\) And, \(f(x)=\frac{x+5}{x+3}\) if \(2\lt x \leq 4\) at \(\mathrm{x}=2\) \(\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}} \frac{x+5}{x+3}=\frac{2+5}{2+3}=\frac{7}{5}\) \(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} \frac{1}{x-1}=\frac{1}{2-1}=1\) \(\therefore \lim _{x \rightarrow 2^{+}} \mathrm{f}(\mathrm{x}) \neq \lim _{\mathrm{x} \rightarrow 2^{-}} \mathrm{f}(\mathrm{x})\) \(\therefore \mathrm{f}(\mathrm{x})\) is discontinuous at \(\mathrm{x}=2\)
MHT CET-2020
Limits, Continuity and Differentiability
79887
If the function \(f(x)=\frac{(1-\cos 5 x) \sin 5 x}{x^{2} \sin 3 x}, x \neq 0\), is continuous at \(x=0\), then \(f(0)=\)
79889
If \(\begin{aligned} f(x) & =6 \beta-3 \alpha x, \text { if }-4 \leq x\lt -2 \\ & =4 x+1, \text { if }-2 \leq x \leq 2\end{aligned}\) is continuous on \([-4,2]\), then \(\alpha+\beta=\)
79886
The points of discontinuity of the function \(\begin{aligned} & f(x)=\frac{1}{x-1} \text { if } 0 \leq x \leq 2 \\ & f(x)=\frac{x+5}{x+3} \text { if } 2\lt x \leq 4\end{aligned}\), in its domain are
1 \(x=2\) only
2 \(x=1, x=2\)
3 \(x=0, x=2\)
4 \(x=4\) only
Explanation:
(B) : Given, \(\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}-1}\) if \(0 \leq \mathrm{x} \leq 2\) at \(x=1=\lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1} \frac{1}{x-1}=\frac{1}{0}=\infty\) \(\therefore \mathrm{f}(\mathrm{x})\) is discontinuous at \(\mathrm{x}=1\) And, \(f(x)=\frac{x+5}{x+3}\) if \(2\lt x \leq 4\) at \(\mathrm{x}=2\) \(\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}} \frac{x+5}{x+3}=\frac{2+5}{2+3}=\frac{7}{5}\) \(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} \frac{1}{x-1}=\frac{1}{2-1}=1\) \(\therefore \lim _{x \rightarrow 2^{+}} \mathrm{f}(\mathrm{x}) \neq \lim _{\mathrm{x} \rightarrow 2^{-}} \mathrm{f}(\mathrm{x})\) \(\therefore \mathrm{f}(\mathrm{x})\) is discontinuous at \(\mathrm{x}=2\)
MHT CET-2020
Limits, Continuity and Differentiability
79887
If the function \(f(x)=\frac{(1-\cos 5 x) \sin 5 x}{x^{2} \sin 3 x}, x \neq 0\), is continuous at \(x=0\), then \(f(0)=\)
79889
If \(\begin{aligned} f(x) & =6 \beta-3 \alpha x, \text { if }-4 \leq x\lt -2 \\ & =4 x+1, \text { if }-2 \leq x \leq 2\end{aligned}\) is continuous on \([-4,2]\), then \(\alpha+\beta=\)
